[tex] y^{2}+xy+x^{2}-7=0 [/tex]
Use the quadratic formula to find:
[tex] y_{1,2}(x) =\frac{-x\pm\sqrt{28-3x^{2}}}{2} [/tex]
(1)
The condition that the point (1,2) should be on the graph of "y" yields:
[tex] y_{1}(1)=\frac{-1+\sqrt{28-3}}{2}=\frac{-1+5}{2}=2 [/tex](2)

[tex] y_{2}(1)=\frac{-1-5}{2}=-3 [/tex] (3)

So you need to chose the "+" sign from the explicitation.
[tex] y(x)=\frac{1}{2}(\sqrt{28-3x^{2}}-x) [/tex]